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2 years ago

"In analyzing distances by apply ing the physics of gravitational forces, an astronomer has obtained the expression

Physics

1 answer:

zavuch27 [327]2 years ago

6 0

Answer:

The value of R is $1.72\times10^{11}\ m$ .

(B) is correct option.

Explanation:

Given that,

In analyzing distances by apply ing the physics of gravitational forces, an astronomer has obtained the expression

$R=\sqrt{\dfrac{1}{(\dfrac{1}{140\times10^{9}})^2-(\dfrac{1}{208\times10^{9}})^2}}$

We need to calculate this for value of R

$R=\sqrt{\dfrac{1}{(\dfrac{1}{140\times10^{9}})^2-(\dfrac{1}{208\times10^{9}})^2}}$

$R=1.89\times10^{11}\ m$

So, The nearest option of the value of R is $1.72\times10^{11}\ m$

Hence, The value of R is $1.72\times10^{11}\ m$ .

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1200 N-m of torque is used to drive a gear (A) of diameter 25 cm, which in turn drives another gear (B) of diameter 52 cm. What

NNADVOKAT [17]

Answer:

2.5kN.m

Explanation:

Torque is directly proportional to pitch diameter

= Ta/Tb= Da/Db

=120/Tb= 0.25/0.5

Tb= 2.469kN.m approx 2.5kN.m

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2 years ago

Read 2 more answers

A horizontal pipe of diameter 0.81 m has a smooth constriction to a section of diameter 0.486 m . The density of oil flowing in

Lerok [7]

Answer:

2.06 m³/s

Explanation:

diameter of pipe, d = 0.81 m

diameter of constriction, d' = 0.486 m

radius, r = 0.405 m

r' = 0.243 m

density of oil, ρ = 821 kg/m³

Pressure in the pipe, P = 7970 N/m²

Pressure at the constriction, P' = 5977.5 N/m²

Let v and v' is the velocity of fluid in the pipe and at the constriction.

By use of the equation of continuity

A x v = A' x v'

r² x v = r'² x v'

0.405 x 0.405 x v = 0.243 x 0.243 x v'

v = 0.36 v' .... (1)

Use of Bernoulli's theorem

$P+\frac{1}{2} \rho v^{2}=P' +\frac{1}{2}\rho'v'^{2}$

7970 + 0.5 x 821 x 0.36 x 0.36 x v'² = 5977.5 + 0.5 x 821 x v'² from (i)

1992.5 = 357.3 v'²

v' = 5.58 m/s

v = 0.36 x 5.58

v = 2 m/s

Rate of flow = A x v = 3.14 x 0.405 x 0.405 x 2 x 2 = 2.06 m³/s

Thus the rate of flow of volume is 2.06 m³/s.

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2 years ago

A projectile is launched at an angle of 60° from the horizontal and at a velocity of

gayaneshka [121]

Answer:

60*12.0= 720 = v/60 * 12.0 squared which is 1,728

Explanation:

Horizontal velocity component: Vx = V * cos(α)

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2 years ago

A force of 150 N accelerates a 25 kg wooden chair across a wood floor at 4.3 m/s2 . How big is the frictional force on the block

solniwko [45]

We can first calculate the net force using the given information.

By Newton's second law, F(net) = ma:

F(net) = 25 * 4.3 = 107.5

We can now calculate the frictional force, f, which is working against the applied force, F(app) (this is why the net force is a bit lower):

f = F(net) - F(app) = 150 - 107.5 = 42.5 N

Now we can calculate the coefficient of friction, u, using the normal force, F(N):

f = uF(n) --> u = f/F(N)
u = 42.5/[25(9.8)]
u = 0.17

4 0

2 years ago

A 0.200-kg mass attached to the end of a spring causes it to stretch 5.0 cm. If another 0.200-kg mass is added to the spring, th

ziro4ka [17]

Answer:

A: 4 times as much

B: 200 N/m

C: 5000 N

D: 84,8 J

Explanation:

In the first question, we have to caculate the constant of the spring with this equation:

$m*g=k*x$

Getting the k:

$k=\frac{m*g}{x} =\frac{0,2[kg]*9,81[\frac{m}{s^{2} } ]}{0,05[m]} =39,24[\frac{N}{m}]$

Then we can calculate how much the spring stretch whith the another mass of 0,2kg:

$x=\frac{m*g}{k} =\frac{0,4[kg]*9,81[\frac{m}{s^{2} } ]}{39,24[\frac{N}{m}]} =0,1[m]\\$

The energy of a spring:

$E=\frac{1}{2}*k*x^{2}$

For the first case:

$E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,05[m])^{2} =0,049 [J]$

For the second case:

$E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,1[m])^{2} =0,0196 [J]$

If you take the relation E2/E1 = 4.

We have the next facts:

x=0,005 m

E = 0,0025 J

Using the energy equation for a spring:

$E=\frac{1}{2}*k*x^{2}$ ⇒ $k=\frac{E*2}{x^{2} } =\frac{0,0025[J]*2}{(0,005[m])^{2} } =200[\frac{N}{m} ]$

The potential energy of the diver will be equal to the kinetic energy in the moment befover hitting the watter.

$E=W*h=500[N]*10[m]=5000[J]$

Watch out the units in this case, the 500 N reffer to the weighs of the diver almost relative to the earth, thats equal to m*g.

The work is equal to the force acting in the direction of the motion. so we have to do the diference beetwen angles to obtain the effective angle where the force is acting: 47-15=32 degree.

The force acting in the direction of the ramp will be the projection of the force in the ramp, equal to F*cos(32). The work will be:

W=F*d=F*cos(32)*d=10N*cos(32)*10m=84,8J

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2 years ago